What is the range of f(x)= 1/(x^2-9) ?

The range of f(x)= 1/(x^2-9) is f(x) 0

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range of f(x)= 1/(x^2-9)

range

f (x) = \frac{1}{x ^{2} - 9}

f (x) \leq - \frac{1}{9} or f (x) > 0

Interval Notation

(- \infty, - \frac{1}{9}] \cup (0, \infty)

Solution steps

Rewrite as

\frac{1}{x ^{2} - 9} = y

Multiply both sides by

x^{2} - 9

1 = y (x^{2} - 9)

The range is the set of y for which the discriminant is greater or equal to zero

Discriminant

1 = y (x^{2} - 9) : 36 y^{2} + 4 y

36 y^{2} + 4 y \geq 0 : y \leq - \frac{1}{9} or y \geq 0

Check if the range interval endpoints are included

y = - \frac{1}{9} :

Included

y = 0 :

Excluded

Therefore the range is

f (x) \leq - \frac{1}{9} or f (x) > 0

h (x) = 2 x - 6

4 x - 3 y = 4

9 x^{3} + 8

f (y) = 2^{x}

f (x) = \frac{x + 2}{1 - x}

What is the range of f(x)= 1/(x^2-9) ?
The range of f(x)= 1/(x^2-9) is f(x)<=-1/9 \lor f(x)>0